Solve $\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$ with initial condition $u(0,y)=0.$ Solve the PDE $\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$ with initial condition $u(0,y)=0.$
I solved the PDE, the solution is $u(x,y)=-y\phi(3x-y)+\frac{x^2y}{2}$, where $\phi$ is a real arbitrary function.
And I am having trouble with the initial condition:
$u(0,y)=-y\phi(-y)=0$.
The issue is the $-y$ factor outside the function $\phi$.
What can I do from here in order to get the complete answer $u(x,y)$?
Any kind of help is very appreciated
 A: I'm not sure how you solved it in the first case so I'll write out a full solution. We use the method of characteristics. Characteristics $(x(s),y(s))$ originating from points $(0,\eta)$ for some $\eta \in \mathbb R$ must obey \begin{align*}
\dot x(s) &= 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x(0) = 0, \\
\dot y(s) &= 3, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y(0) = \eta, \\
\dot z(s) &= x(s)y(s), \,\,\,\, z(0) = 0, 
\end{align*} where $z(s) = u(x(s),y(s)$. Solving the first two equations gives $x(s) = s$, $y(s) = 3s + \eta$ and thus $$\dot z(s) = 3s^2 + \eta s \,\,\,\, \implies z(s) - \underbrace{z(0)}_{\,\,\,\,\,= 0} = s^3 + \frac{\eta s^2}{2}.$$ Now if we can invert the relationships (i.e., solve for $(s,\eta)$ in terms of $(x,y)$) we will have $$u(x,y) = z(s(x,y); \eta(x,y)).$$ This is very easy to do: we have $s = x$ and $\eta = y - 3x$. Thus $$\boxed{u(x,y) = x^3 + \frac{x^2(y-3x)}{2}= \frac{1}{2}(x^2y - x^3)}.$$ Indeed, we can check the answer: it's clear that $u(0,y) = 0$ and $$\frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = xy - \frac{3}{2}x^2 + 3\cdot\frac{1}{2}x^2 = xy.$$
A: You were close to the solution you just did I guess some mistakes
$$\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$$
The differential equations are
$$\frac {dx}1=\frac {dy}3=\frac {dz}{xy}$$
$$
\begin{cases}
\int 3dx= \int  dy \implies y=3x+K_1 \\
xydy=dz \implies \int x(3x+K_1)dx=z+K_2 \\ 
\implies K_2=x^3+\frac {x^2K_1}2-z
\end{cases}
$$
Therefore
$$\phi((y-3x)=\frac  {-x^3}2+\frac {x^2y}2-z \implies u(x,y)=-\phi((y-3x)-\frac  {x^3}2+\frac {x^2y}2$$
Initital condition $u(0,y)=0 \implies \phi(y-3x)=0$
Therefore the solution is
$$\boxed{ u(x,y)=\frac {x^2}2( y-x)}$$
